If not, you are told whether or not the prize is in an adjacent door. (note: if you pick door A, and they say its adjacent, this means it HAS to be door B)

You then get a second guess.

What are the chances of winning the game if you pick door A or F? What about picking B, C, D, or E?

Apr 28th 2010, 08:17 PM

CaptainBlack

Quote:

Originally Posted by shenanigans87

Here's the situation. You have 6 six doors.

A B C D E F

Behind one of them is a prize.

You get to choose a door.

If it has the prize, you are told so and win immediately.

If not, you are told whether or not the prize is in an adjacent door. (note: if you pick door A, and they say its adjacent, this means it HAS to be door B)

You then get a second guess.

What are the chances of winning the game if you pick door A or F? What about picking B, C, D, or E?

If you pick door A your chance of winning is 1/3 since you win if it was in fact behind doors A or B, same reasoning applies to door F.

CB

Apr 28th 2010, 08:51 PM

shenanigans87

Close... but that is incorrect. No matter what door you pick, you have a 50% chance of winning, assuming you use common sense regarding the adjacent hints. For example... pretend your strategy is as follows: you always choose door A. If wrong and adjacent you know its B. If not adjacent you always guess C. Thus, if the prize was behind doors A B or C you will be guaranteed a win. And there's a 3/6 or 50% chance its behind doors A B or C.

Apr 28th 2010, 10:55 PM

CaptainBlack

Quote:

Originally Posted by shenanigans87

Close... but that is incorrect. No matter what door you pick, you have a 50% chance of winning, assuming you use common sense regarding the adjacent hints. For example... pretend your strategy is as follows: you always choose door A. If wrong and adjacent you know its B. If not adjacent you always guess C. Thus, if the prize was behind doors A B or C you will be guaranteed a win. And there's a 3/6 or 50% chance its behind doors A B or C.

If you choose door A there is no strategy involved, you win if the prize is behind door A or door B, which is 1/3.

The same if you pick door F

The above is what the first part of the question asks for.

CB

Apr 29th 2010, 06:47 AM

shenanigans87

First of all, this theory has already been proven both by many stats majors and thinkers alike. Once you understand it, it's really not that complicated of a problem. Secondly, I made a computer simulation of this scenario, and tested it for both method A and method B 500,000 times. At any given moment from 1,000 trials to 500,000 trials, the ratio was within 0.01% of 50/50

Method A: You have a 1/6 chance of winning right off the bat. However, if you don't, there is a 1/5 chance that the prize is next to you, and a 4/5 chance that the prize is far from you, in which case you have a 1/4 chance of guessing correctly. =)

Method B: You have a 1/6 chance of winning from the start. If you are wrong, and the prize is adjacent, which will happen 2/5 of the time, you have a 1/2 chance of guessing right. If the prize is far, which will happen 3/5 of the time, you have a 1/3 chance of guessing right.

There you have it. I was initially very frustrated with this problem as well, but once I figured it out and was then able to prove the theory by running a relatively infinite amount of trials, it made sense.